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To proceed by induction, assume that inequality (3.15) holds for step (i 1) of the path, i > 1: {SCi 1 } + |Ci 1 Ti 1 | [STi 1 ] (3.16)Create Barcode In Visual Studio .NETUsing Barcode encoder for VS .NET Control to generate, create bar code image in VS .NET applications.Each step of the robot s path takes place in one of two ways: either Ci 1 = Ti 1 or Ci 1 = Ti 1 . The latter case takes place when the robot moves along the locally convex part of an obstacle boundary; the former case comprises all the remaining situations. Consider the rst case, Ci 1 = Ti 1 . Here the robot will take a step of length |Ci 1 Ci | along a straight line toward Ti 1 ; Eq. (3.16) can thus be rewritten as {SCi 1 } + |Ci 1 Ci | + |Ci Ti 1 | [STi 1 ] In (3.17), the rst two terms form {SCi }, and so {SCi } + |Ci Ti 1 | [STi 1 ] (3.18) (3.17)Barcode Decoder In VS .NETUsing Barcode recognizer for VS .NET Control to read, scan read, scan image in .NET applications.MOTION PLANNING FOR A MOBILE ROBOT Paint PDF417 In C#Using Barcode creation for Visual Studio .NET Control to generate, create PDF-417 2d barcode image in Visual Studio .NET applications.At point Ci the robot will de ne the next intermediate target, Ti . Now add to (3.18) the obvious inequality |Ti 1 Ti | [Ti 1 Ti ]: {SCi } + |Ci Ti 1 | + |Ti 1 Ti | [STi 1 ] + [Ti 1 Ti ] = [STi ] By the Triangle Inequality, we have |Ci Ti | |Ci Ti 1 | + |Ti 1 Ti | Therefore, it follows from (3.19) and (3.20) that {SCi } + |Ci Ti | [STi ] (3.21) (3.20) (3.19)Create PDF417 In .NET FrameworkUsing Barcode creation for ASP.NET Control to generate, create PDF-417 2d barcode image in ASP.NET applications.which proves (3.15). Consider now the second case, Ci 1 = Ti 1 . Here the robot takes a step of length (Ci 1 Ci ) along the obstacle boundary (the Bug2 path, [Ci 1 Ci ]). Equation (3.16) becomes {SCi 1 } + [Ci 1 Ci ] [SCi 1 ] + [Ci 1 Ci ] (3.22)Drawing PDF-417 2d Barcode In VB.NETUsing Barcode creator for Visual Studio .NET Control to generate, create PDF 417 image in .NET framework applications.where the left-hand side amounts to {SCi } and the right-hand side to [SCi ]. At point Ci , the robot will de ne the next intermediate target, Ti . Since |Ci Ti | [Ci Ti ], inequality (3.22) can be written as {SCi } + |Ci Ti | [SCi ] + [Ci Ti ] = [STi ] (3.23)EAN-13 Drawer In Visual Studio .NETUsing Barcode creator for .NET framework Control to generate, create EAN13 image in VS .NET applications.which, again, produces (3.15). Since, by the algorithm s design, at some nite i, Ci = T , then {ST } [ST ] which completes the proof. Q.E.D. One can also see from Figure 3.15 that when rv goes to in nity, algorithm VisBug-21 will generate locally optimal paths, in the following sense. Take two obstacles or two parts of the same obstacle, k and k + 1, that are visited by the robot, in this order. During the robot s passing around obstacle k, once a point on obstacle k + 1 is identi ed as the next intermediate target, the gap between k and k + 1 will be traversed along the straight line, which presents the locally shortest path. When de ning its intermediate targets, algorithm VisBug-21 could sometimes use points on the M-line that are not necessarily contiguous to the prior intermediate targets. This would result in a more ef cient use of robot s vision: By cutting corners, the robot would be able to skip some obstacles that intersect the M-line and that it would otherwise have to traverse. 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